Equivalence relation

Step 1: The greatest integer part of (2.4) is (2). Step 2: Real numbers with greatest integer part (2) are greater than or equal to (2) and less than (3). Step 3: In such questions, check the end points carefully, so the class is ([2,3)).

Step 1: The remainder of (-1) modulo (4) is written as (3). Step 2: So all integers related to (-1) leave remainder (3) on division by (4). Step 3: Even for negative numbers, write the remainder in (0,1,2,3) for exam clarity.

Step 1: The set ({1,2}) has (2) elements. Step 2: Its equivalence class contains all subsets of (A) having exactly (2) elements. Step 3: The number of such subsets is (\binom{4}{2}=6), so order should not be counted.

Step 1: For every real number, (a^3=a^3), so the relation is reflexive. Step 2: Equality of cubes works in both directions and is transitive. Step 3: Over real numbers, (a^3=b^3) implies (a=b), so every equivalence class is singleton.

Step 1: A line has the same direction as itself, so the relation is reflexive. Step 2: Same direction is mutual and remains consistent through a third line. Step 3: Hence each equivalence class contains lines with one fixed direction, not necessarily lines through one point.

Step 1: Division by (3) creates three equivalence classes. Step 2: Each class has (3) elements, so each contributes (3^2=9) ordered pairs. Step 3: Total pairs are (9+9+9=27); pairs across different classes are not counted.

Step 1: For every (a), (\sin a=\sin a), so the relation is reflexive. Step 2: If two sine values are equal, equality also holds in the reverse direction. Step 3: A relation based on equal function values is transitive too, so it is an equivalence relation.

Step 1: (|a-b|) being even means (a) and (b) have the same parity. Step 2: Since (4) is even, its related elements must be even. Step 3: In the given set, the even elements are (2) and (4), so the class is ({2,4}).

Step 1: For any polynomial, (p(0)=p(0)), so reflexivity holds. Step 2: If (p(0)=q(0)), then (q(0)=p(0)) also holds. Step 3: Having the same value at zero passes through a third polynomial, so the relation is equivalence.

Step 1: All ((a,a)) pairs are present, so reflexivity holds. Step 2: Symmetry is also visible from the given pairs. Step 3: But from ((2,1)) and ((1,3)), transitivity requires ((2,3)), which is missing; hence it is not an equivalence relation.

Step 1: In an equivalence relation, all ordered pairs inside each block are included. Step 2: The first block gives (2^2=4) pairs and the second gives (3^2=9) pairs. Step 3: Total pairs are (4+9=13); (5^2) pairs would occur only if there were one class.

Step 1: (ab>0) means (a) and (b) are either both positive or both negative. Step 2: So all positive integers form one class and all negative integers form another. Step 3: Zero is not included in the set, so no separate zero class is formed.

Step 1: The radius of any circle is equal to itself. Step 2: If the first circle has the same radius as the second, the reverse is also true. Step 3: Equality of radius is transitive, so this is an equivalence relation.

Step 1: (a+b) is even only when (a) and (b) are both even or both odd. Step 2: In the given set, the odd elements are (1,3,5) and the even elements are (2,4,6). Step 3: Hence the relation forms two classes based on parity.

Step 1: For reflexivity, every ((a,b)) must satisfy ((a,b)S(a,b)). Step 2: This would require (a+a=b+b), meaning (a=b). Step 3: Not every ordered pair has (a=b), for example ((1,2)), so the relation is not equivalence.

Step 1: (12=2^2\cdot 3), so its set of prime factors is ({2,3}). Step 2: (18=2\cdot 3^2) also has the prime factor set ({2,3}). Step 3: Powers may differ, but this relation checks only the set of prime factors.

Step 1: Equivalence classes are either completely disjoint or exactly the same. Step 2: If two classes have a common element, that element connects both classes. Step 3: Therefore the two classes cannot remain distinct, so (B_1=B_2).

Step 1: In (A\times A), every element of (A) is related to every element of (A). Step 2: Therefore no two elements can lie in different classes. Step 3: This is the universal relation, and its single equivalence class is the whole set (A).

Step 1: (3) itself is a multiple of (3). Step 2: Its equivalence class contains exactly those elements that are multiples of (3). Step 3: In the given set, the multiples of (3) are (3) and (6), so the class is ({3,6}).

Step 1: A number (x) is related to (1.7) only when (x-1.7) is an integer. Step 2: This means (x=1.7+n), where (n\in Z). Step 3: Writing just one interval is wrong because the class contains numbers at integer distances.

Step 1: (\min(5,3)=3). Step 2: The elements for which (\min(a,3)=3) are (3,4,5). Step 3: In a relation made by equal function values, all elements with the same value lie in one class.

Step 1: The main property of equivalence classes is that two classes are either equal or disjoint. Step 2: Here ([a]\neq[b]), so they are not equal. Step 3: Therefore their intersection must be empty.

Step 1: (4) leaves remainder (0) when divided by (2). Step 2: (7) leaves remainder (1) when divided by (2). Step 3: The remainders are different, so (4) and (7) are not related and ((4,7)\notin R).

Step 1: Every student has the same date of birth as himself or herself. Step 2: If the first student has the same date as the second, the second has the same date as the first. Step 3: If two students share the same date with a third student, they share it with each other too.

Step 1: To be an equivalence relation, at least every ((a,a)) pair is needed. Step 2: Taking only these reflexive pairs still preserves symmetry and transitivity. Step 3: Therefore the equality relation is the smallest equivalence relation.